# Tag Info

## Hot answers tagged pde

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what are the natural spaces Depends on the task at hand. Function spaces are tools: a hammer is a natural tool for one task, a screwdriver for another. I believe one usually considers $u\in C^2\cap L^2$ and $g\in L^2$ I would not call this usual. The heat equation is of first order in $t$; why ask for two derivatives with respect to $t$? And ...

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I'll use notation $\Delta u$ instead of $-\mu$, because it fits better in the formulas. Take a smooth domain $G$ such that $\operatorname{supp} \Delta u\subset G \Subset \Omega$. Using a smooth cutoff function, write $\psi$ as $\psi_1+\psi_2$ where $\psi_1$ has compact support in $\Omega$ and $\psi_2\equiv 0$ on $\overline{G}$. Apply Green's identity to ...

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Some extra references that might be useful (I have at least found them useful on related topics): W. Hayman: Subharmonic Functions Volume 2 - in particular, section 7.3, geometric estimates for capacity. Garnett, Marshall: Harmonic Measure Doob: Classical Potential Theory and its Probabilistic Counterpart - see Part I, chapter XIII.

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Take your $R_n(r) = Ar^n + Br^{-n}$. We want this to hold for all possible values on $r$, however we still want our solution to be physically plausible. This means that it can not run away to infinite, i.e. we need to care about when $r=0$. Setting $R_n(r) = c_nr^{|n|}$ fixes this problem, with $c_n = A,B$ depending on $n$.

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Why are PDE's classified in this way where it relates to conic sections? They are not. The parallel with conic sections is an artifact of second-order PDE in two dimensions. It is not a classification of PDE in general, as one quickly discovers when encountering higher order equations and higher dimensions. The properties recognized in two dimensions ...

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I'm separating the variables using $u(x,y)=X(x)Y(y)$ and trying to to incorporate the b.c.: \begin{align} X'' - \lambda^2 X &= 0\\ X(x=0) &= \sin(y)\\ X(x=1) &= e \sin(y) \end{align} Your solution is off the rails already here. $X$ is a function of $x$ only, that's the idea of separation of variables. Asking for $X(x=0)=\sin y$ makes ...

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Yes, this is true, but a complete proof requires a fair amount of work. I have a note that Wloka's Partial Differential Equations, Section 25, is a good place to look, but I don't have a copy handy to check right now. But very roughly speaking, the outline is like this. Let $\mathcal{H}$ be your space of $L^2(0,T; H^1)$ functions with a weak derivative in ...

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If $u$ has compact support in $\mathbb R^n$... This is what happens when people try to pack too much information into too few words. They meant to say If $\Omega=\mathbb R^n$ and $u$ has compact support... Notice that there is no $\Omega$ under the integral on the next line, indicating they are thinking of $\mathbb R^n$. Why $\mathbb R^n$, ...

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The 2-D heat equation is: $$\frac{\partial T}{\partial t} - \alpha \Delta T = 0$$ And for steady state we know that $\partial T / \partial t = 0$ which simplifies our equation to $$\Delta T = 0$$ which in polar coordinates is $$\frac{1}{r}\frac{\partial}{\partial r}\left(\frac{\partial T}{\partial ... 1 I think that as long as everything is well defined (for example, being in the Schwartz space), then the result is what you think it should be. For example,$$F\left(\frac{\partial^2}{\partial x \partial y}\right) u(x,y,t) = (ix)(iy) \hat{u}(x,y,t)$$(assuming you are doing the transform on x and y). This is stated in Stein's book on Fourier Analysis ... 1 What happens depends on what variable(s) you are applying the Fourier transform to; if we suppose we are making the Fourier transform with regards to x \rightsquigarrow \xi, then if we are using the one-dimensional convention that$$ \hat{f}(\xi) = \int^{\infty}_{-\infty}f(x)e^{-i\xi x} dx$$so in higher dimensions we generalise to$$ \hat{f}(\xi,y,t) ...

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The usual Fourier series method shows that a solution is given by $$u(x,t) = \sum_{n=1,3,5,\dots} \frac{2}{\pi n} e^{-n^2 \pi^2 t} \sin(n \pi x).$$ You can check that this satisfies the boundary and initial conditions in the following senses: it extends continuously to $([0,1] \times (0,\infty)) \cup ((0,1) \times \{0\})$, and the extension equals 0 on ...

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Suggestion: start with a symmetric operator $C$ that is not essentially self adjoint, and $A$ with ${\mathcal D}(A) \subset {\mathcal D}(C)$ that is self adjoint. Take $B = C - A$.

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If $\Delta u$ is a signed measure, it can be decomposed into a positive and negative parts. Consequently, $u$ can be be written as a difference of two subharmonic functions. Such $u$ is called $\delta$-subharmonic. There are $W^{1,1}$ functions that are not $\delta$-subharmonic. I'll give an example in one dimension: $u(x)=\int_0^x W(t)\,dt$ where $W$ is ...

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Taking $\partial_x$ on the first equation and $\partial_y$ on the second equation we get $${\partial^3 H_y \over \partial x^2 \partial y} = {\partial^3 H_x \over \partial x\partial y^2} + \mu {\partial H_x \over \partial x}$$ {\partial^3 H_x \over \partial x \partial y^2} = {\partial^3 H_y \over \partial x^2\partial y} + \mu {\partial H_y \over \partial ...

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